If $t$ is any real number, then I can find a strictly increasing sequence $(t_n)$ of real numbers converging to $t$. E.g. $t_n = t - \frac{1}{n}$ would do.

Does this generalize to arbitrary partially ordered sets $(P,\leq)$? If $t$ is an element in $P$, then is there a sequence $(t_n)$ in $P$ such that it is strictly increasing (i.e., $t_n\leq t_{n+1}$ and $t_n \neq t$ for every $n$) and converges to $t$ (i.e., for every $s$ in $P$ there is an integer $n$ such that $t_n\leq s$)?

My guess is that the answer is no, that it fails for sufficiently large partially ordered sets, but I do not know of a counterexample. I looked a bit into ordinal numbers, thinking that I might find an answer / counterexample there, but I am not at all familiar with that area. Any help would be appreciated!


1 Answer 1


Your guess is correct, and you were looking in the right place. Let $P$ be the set of ordinals $\alpha$ such that $\alpha\le\omega_1$, $\omega_1$ being the first uncountable ordinal. Suppose that $\langle\alpha_n:n\in\Bbb N\rangle$ is a strictly increasing sequence of ordinals less than $\omega_1$. Each $\alpha_n$ has only countably many predecessors, and each ordinal is by definition the set of its predecessors, so

$$\beta=\bigcup_{n\in\Bbb N}\alpha_n$$

is the union of countably many countable sets and hence is countable. The set $\beta$ is in fact an ordinal, the supremum of the ordinals $\alpha_n$, and it is strictly less than $\omega_1$, since $\omega_1$ has uncountably many predecessors. Thus, no sequence of smaller ordinals converges to $\omega_1$, the top element of $P$.

  • $\begingroup$ Do you mean $\alpha < \omega_1$ rather then $\alpha \le \omega_1$ ? $\endgroup$
    – ak87
    Apr 11, 2018 at 14:19

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